91影视

When a chemical process reaches equilibrium, the equilibrium constant (typically indicated by the symbol) offers information on the relationship between the products and reactants.

$\text{0}$The information provided is 鈥�

$\begin{array}{c}{\text{K}}_{\text{a}}{\text{=7.1脳10}}^{\text{-4}}\\ {\text{[HNO}}_{\text{2}}\text{]=0.60M}\end{array}$

The reaction for the dissociation of${\text{HNO}}_{\text{2}}$is 鈥�

${\text{HNO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}鈬�{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+NO}}_{\text{2}}^{\text{-}}$

Construct the ICE table to obtain the equation for ${\text{K}}_{\text{a}}$.

Write the expression for the equilibrium constant of the reaction in terms of concentration 鈥�

$\begin{array}{l}{\text{K}}_a\text{=}\frac{\text{Products}}{\text{Reactants}}\\ {\text{K}}_a\text{=}\frac{{\text{[NO}}_{\text{2}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\end{array}$

Substitute the equilibrium equations from the reaction table to solve for 鈥�

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[NO}}_{\text{2}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{{}_{\text{2}}}}{\text{0.60-x}}\end{array}$

Substitute the equilibrium equations from the reaction table to solve for 鈥�

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[NO}}_{\text{2}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{{}_{\text{2}}}}{\text{0.60-x}}\end{array}$

It is known that that ${\text{x=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]=[NO}}_{\text{2}}^{\text{-}}\text{]}$. Since ${\text{HNO}}_{\text{2}}$is a weak acid, it鈥檚 ${\text{K}}_{\text{a}}$must be very small. So, assume that the $x$has no effect on $\text{0.60M}$in the denominator. Then substitute the ${\text{K}}_{\text{a}}$to solve for $x$.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.60}}\\ {\text{x}}^{\text{2}}\text{=}\left({\text{K}}_{\text{a}}\right)\text{(0.60)}\\ \text{x=}\sqrt{\left({\text{K}}_{\text{a}}\right)\text{(0.60)}}\\ \text{=}\sqrt{\left({\text{7.1脳10}}^{\text{-4}}\right)\text{(0.60)}}\\ \text{x=0.021}\\ {\text{=2.1脳10}}^{\text{-2}}\end{array}$

Since${\text{x=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]=[NO}}_{\text{2}}^{\text{-}}\text{]}$then ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]=[NO}}_{\text{2}}^{\text{-}}{\text{闭=2.1脳10}}^{\text{-2}}$.

Now, solve for ${\text{[OH}}^{\text{-}}\text{]}$using ${\text{K}}_{\text{w}}{\text{=1.0脳10}}^{\text{-14}}$.

$\begin{array}{c}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \text{=}\frac{{\text{1.0脳10}}^{\text{-14}}}{{\text{2.1脳10}}^{\text{-2}}}\\ \left[{\text{OH}}^{\text{-}}\right]{\text{=4.8脳10}}^{\text{-13}}\end{array}$

Therefore, the values of concentrations of${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{],[NO}}_{\text{2}}^{\text{-}}\text{]}$ and${\text{[OH}}^{\text{-}}\text{]}$ are${\text{2.1脳10}}^{\text{-2}}{\text{,2.1脳10}}^{\text{-2}}$ and${\text{4.8脳10}}^{\text{-13}}$ respectively.